You have some number of sticks with positive integer lengths. These lengths are given as an array sticks
, where sticks[i]
is the length of the ith
stick.
You can connect any two sticks of lengths x
and y
into one stick by paying a cost of x + y
. You must connect all the sticks until there is only one stick remaining.
Return the minimum cost of connecting all the given sticks into one stick in this way.
Example 1:
Input: sticks = [2,4,3] Output: 14 Explanation: You start with sticks = [2,4,3]. 1. Combine sticks 2 and 3 for a cost of 2 + 3 = 5. Now you have sticks = [5,4]. 2. Combine sticks 5 and 4 for a cost of 5 + 4 = 9. Now you have sticks = [9]. There is only one stick left, so you are done. The total cost is 5 + 9 = 14.
Example 2:
Input: sticks = [1,8,3,5] Output: 30 Explanation: You start with sticks = [1,8,3,5]. 1. Combine sticks 1 and 3 for a cost of 1 + 3 = 4. Now you have sticks = [4,8,5]. 2. Combine sticks 4 and 5 for a cost of 4 + 5 = 9. Now you have sticks = [9,8]. 3. Combine sticks 9 and 8 for a cost of 9 + 8 = 17. Now you have sticks = [17]. There is only one stick left, so you are done. The total cost is 4 + 9 + 17 = 30.
Example 3:
Input: sticks = [5] Output: 0 Explanation: There is only one stick, so you don't need to do anything. The total cost is 0.
Constraints:
1 <= sticks.length <= 104
1 <= sticks[i] <= 104
连接棒材的最低费用。
为了装修新房,你需要加工一些长度为正整数的棒材 。棒材以数组 sticks 的形式给出,其中 sticks[i] 是第 i 根棒材的长度。
如果要将长度分别为 x 和 y 的两根棒材连接在一起,你需要支付 x + y 的费用。 由于施工需要,你必须将所有棒材连接成一根。
返回你把所有棒材 sticks 连成一根所需要的最低费用。注意你可以任意选择棒材连接的顺序。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/minimum-cost-to-connect-sticks
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思路是贪心。题目指出连接棒材的费用是两个棒材的长度之和,同时因为题目最终要达到的目的是把 input 中所有的棒材连接在一起,所以做法应该是优先找长度最短的两根棒材连接,这样连接的费用始终是input中长度最短的两根棒材的长度之和。
时间O(nlogn)
空间O(n)
Java实现
1 class Solution { 2 public int connectSticks(int[] sticks) { 3 PriorityQueue<Integer> queue = new PriorityQueue<>(); 4 for (int s : sticks) { 5 queue.offer(s); 6 } 7 int sum = 0; 8 while (queue.size() > 1) { 9 int first = queue.poll(); 10 int second = queue.poll(); 11 sum += first + second; 12 queue.offer(first + second); 13 } 14 return sum; 15 } 16 }